Note: See update below.

Consider the Majorana Lagrangian

$$\mathcal{L}=-\psi ^{\mathrm{T}}\mathrm{i}% \gamma ^{0}\left( \gamma ^{\rho }\partial _{\rho }+m\right) \psi ,\tag{1}$$

where $% \psi \in \mathbb{R}^{4}$ is Grassmann-valued, and where $\gamma ^{\rho }\in \mathrm{M}_{4}\left( \mathbb{R}\right) $ is the irreducible, real-valued representation of $\mathrm{Cliff}\left( 3,1\right) $. The generalized coordinates $\psi ^{a}=\psi ^{a}\left( \mathbf{x}\right) $ and their conjugate momenta $\pi _{a}\left( \mathbf{x}\right) \equiv \pi _{a}^{L}\left( \mathbf{x% }\right) =-\pi _{a}^{R}\left( \mathbf{x}\right) $, where

\begin{eqnarray*} \pi _{a}^{L} &\equiv &\frac{\partial _{L}\mathcal{L}}{\partial \dot{\psi}^{a}% }\equiv \frac{\overrightarrow{\partial }}{\partial \dot{\psi}^{a}}\mathcal{L}% =+\mathrm{i}\left( \psi ^{\mathrm{T}}\right) _{a}=+\mathrm{i}\psi _{a}, \\ \pi _{a}^{R} &\equiv &\frac{\partial _{R}\mathcal{L}}{\partial \dot{\psi}^{a}% }\equiv \mathcal{L}\frac{\overleftarrow{\partial }}{\partial \dot{\psi}^{a}}% =-\mathrm{i}\left( \psi ^{\mathrm{T}}\right) _{a}=-\mathrm{i}\psi _{a},\tag{2} \end{eqnarray*}

following answer by Qmechanic, are obviously not independent, but constrained by $$ 0=\chi _{a \mathbf{x}} \equiv \psi _{a}\left( \mathbf{x}\right) +\mathrm{i}\pi _{a}\left( \mathbf{x}\right) .\tag{3} $$ The Poisson brackets of these constraints with themselves are given by

\begin{eqnarray*} C_{a \mathbf{x},b \mathbf{y}} &\equiv &\left\{ \chi _{a \mathbf{x}},\chi _{b \mathbf{y}} \right\} _{\text{P}} \\ &=&\mathrm{i}\left\{ \psi _{a}\left( \mathbf{x}\right) ,\pi _{b}\left( \mathbf{y}\right) \right\} _{\text{P}}+\mathrm{i}\left\{ \pi _{a}\left( \mathbf{x}\right) ,\psi _{b}\left( \mathbf{y}\right) \right\} _{\text{P}} \\ &=&-2\mathrm{i}\delta _{ab}\delta ^{\left( 3\right) }\left( \mathbf{x}-% \mathbf{y}\right) ,\tag{4} \end{eqnarray*}


\begin{eqnarray*} \left\{ F,G\right\} _{\text{P}} &=&\int d^{3}x\left[ \frac{\delta _{R}F}{ \delta \psi ^{a}\left( \mathbf{x}\right) }\frac{\delta _{L}G}{\delta \pi _{a}^{R}\left( \mathbf{x}\right) }-\frac{\delta _{R}F}{\delta \pi _{a}^{L}\left( \mathbf{x}\right) }\frac{\delta _{L}G}{\delta \psi ^{a}\left( \mathbf{x}\right) }\right] \\ &\equiv &-\int d^{3}x\left[ \frac{\delta _{R}F}{\delta \psi ^{a}\left( \mathbf{x}\right) }\frac{\delta _{L}G}{\delta \pi _{a}\left( \mathbf{x} \right) }+\frac{\delta _{R}F}{\delta \pi _{a}\left( \mathbf{x}\right) }\frac{ \delta _{L}G}{\delta \psi ^{a}\left( \mathbf{x}\right) }\right] ,\tag{5} \end{eqnarray*}

following again answer by Qmechanic. The matrix $C_{a \mathbf{x},b \mathbf{y}}$ being invertible,

$$ \left( C^{-1}\right) _{a\mathbf{x},b\mathbf{y}}=\frac{\mathrm{i}}{2}\delta _{ab}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{y}\right),\tag{6} $$

implies, in the terminology of Dirac, that these constraints are second class. The Dirac bracket is thus given by

\begin{eqnarray*} \left\{ F,G\right\} _{\text{D}} &\equiv &\left\{ F,G\right\} _{\text{P} }-\int d^{3}x\int d^{3}y\left\{ F,\chi ^{a\mathbf{x}}\right\} _{\text{P} }\left( C^{-1}\right) _{a\mathbf{x},b\mathbf{y}}\left\{ \chi ^{b\mathbf{y} },G\right\} _{\text{P}} \\ &=&\left\{ F,G\right\} _{\text{P}}-\frac{\mathrm{i}}{2}\int d^{3}x\left\{ F,\chi ^{a\mathbf{x}}\right\} _{\text{P}}\left\{ \chi _{a\mathbf{x} },G\right\} _{\text{P}},\tag{7} \end{eqnarray*}

so that

\begin{eqnarray*} \left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi ^{b}\left( \mathbf{y}\right) \right\} _{\text{D}} &=&\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi ^{b}\left( \mathbf{y}\right) \right\} _{\text{P}}-\frac{\mathrm{i}}{2}\int d^{3}z\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\chi ^{c\mathbf{z}}\right\} _{\text{P}}\left\{ \chi _{c\mathbf{z}},\psi ^{b}\left( \mathbf{y}\right) \right\} _{\text{P}} \\ &=&\frac{\mathrm{i}}{2}\int d^{3}z\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\pi ^{c}\left( \mathbf{z}\right) \right\} _{\text{P}}\left\{ \pi _{c}\left( \mathbf{z}\right) ,\psi ^{b}\left( \mathbf{y}\right) \right\} _{\text{P}} \\ &=&\frac{\mathrm{i}}{2}\int d^{3}z\delta ^{ac}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{z}\right) \delta _{c}^{b}\delta ^{\left( 3\right) }\left( \mathbf{z}-\mathbf{y}\right) \\ &=&\frac{\mathrm{i}}{2}\delta ^{ab}\delta ^{\left( 3\right) }\left( \mathbf{x }-\mathbf{y}\right) .\tag{8} \end{eqnarray*}

Following the 'quantum bracket = $\mathrm{i} \times$ Poisson bracket'-rule, the quantum bracket is thus supposedly given by

$$ \left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi ^{b}\left( \mathbf{y}\right) \right\} =-\frac{1}{2}\delta ^{ab}\delta ^{\left( 3\right) }\left( \mathbf{x} -\mathbf{y}\right) .\tag{9} $$

Can that really be correct? The minus sign looks wrong. And what about the factor of 1/2? But before turning attention to these details, perhaps I should start by asking whether I have made some conceptual errors above. Have I simply misunderstood how to go about quantizing constrained systems?

Update: Now the penny finally dropped. Almost embarrassingly, the wrong sign boils down to not having taken into account the minus sign in $\gamma _{0}^{2}=-1_{4}$ in the calculation of the conjugate momenta. Taking proper care of this, the sign issue evaporates. (Note that in order to preserve the history of this posting, the material above has not been edited accordingly.) Thus to me only remains now the puzzlement over the factor of 1/2, see my comment below.


1 Answer 1


Comments to the question (v9):

  1. If we ignore the overall normalization, then OP correctly applies the Dirac-Bergmann$^1$ method, which leads to second-class constraints.$^2$

  2. Normally the Majorana Lagrangian (1) is defined with a factor $\frac{1}{2}$ in front. Then there will be no factor $\frac{1}{2}$ in the anti-commutator relation (9), see e.g. Ref. 2.

  3. As for the overall sign of the Lagrangian (1), one should chose consistent sign conventions. In particular, the Hamiltonian should be bounded from below. See e.g. Ref. 2 for a consistent choice.


  1. S. Weinberg, Quantum Theory of Fields, Vol. 1; Section 7.6.

  2. M. Srednicki, QFT, Chapter 37. A prepublication draft PDF file is available here.


$^1$ OP mentions in a comment that he (partially) follows Ref. 1. Note that Ref. 1 does not explain sign conventions for Grassmann-odd fields.

$^2$ For a similar calculation with fermionic second-class constraints, see e.g. my Phys.SE answer here (NB: not Majorana). Instead of the Dirac-Bergmann method, one can use the Faddeev-Jackiw method, cf. e.g. my Phys.SE answer here (NB: not Majorana).

  • $\begingroup$ Thanks for your answer. I am aware of the book by Srednicki, and have been browsing it, but it is not really of interest to me in this connection because it does not treat how to quantize constrained systems; as he writes p. 238: "... requires new formalism for the quantization of constrained systems." In my writings above, I have been adapting the material of Section 7.6, Constraints and Dirac Brackets, of Weinberg's The Quantum Theory of Fields. $\endgroup$ Commented Jun 19, 2015 at 11:02
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Jun 19, 2015 at 11:40
  • $\begingroup$ Thanks for your update, some details of which I am still looking into. I am happy to hear that my calculations are at least conceptually sound. The supposed extra factor of 1/2 in the Majorana Lagrangian puzzles me, though, because in the Majorana representation, the Dirac Lagrangian decomposes into the sum of two independent Lagrangians of the form (1). $\endgroup$ Commented Jun 21, 2015 at 6:15

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